Answer:
#_time = 7.5 10⁴ s
Explanation:
In order for the astronaut to be younger than the people on earth, it follows that the speed of light has a constant speed in vacuum (c = 3 108 m / s), therefore with the expressions of special relativity we have.
t = 
where t_p is the person's own time in an immobile reference frame,
let's calculate
we assume that the speed of the space station is constant
t_ = 0.99998666657 s
therefore the time change is
Δt = t - t_p
Δt = 1 - 0.9998666657
Δt = 1.3333 10⁻⁵ s
this is the delay in each second, therefore we can use a direct rule of proportions. If Δt was delayed every second, how much second (#_time) is needed for a total delay of Δt = 1 s
#_time = 1 / Δt
#_time =
#_time = 7.5 10⁴ s
Answer:
Bok.............................
decreased 5 times
Explanation: if the force increases 5 times between them would decrease 5 times
Answer:
(a) Jx = -1.14Ns, Jy = 110×3×10-³ = 0.330Ns (b) V = (0m/s)ı^−(1.79m/s)ȷ^
Explanation:
Given
W = 0.56N = mg
m = 0.56/g = 0.56/9.8 = 0.057kg
t = 3.00ms = 3.00×10-³s
Impulse is a vector quantity so we would treat it as such
We have been given the force and velocity in their component forms so to get the impulse from these quantities, we pick the respective component for the quantity we want to calculate and do the necessary calculation. The masses are scalar quantities and so do not affect the signs used in the calculations whether positive or negative. So we have that
u = (20.0m/s)ı^−(4.0m/s)ȷ^
ux = 20m/s
uy = – 4.0m/s
F = – (380N)ı^+(110N)ȷ^
Fx = –380N
Fy = 110N
J = impulse = force × time = F×t
So Jx = Fx ×t
Jy = Fy×t
Jx = –380×3×10-³ = -1.14Ns
Jy = 110×3×10-³ = 0.330Ns
Impulse also equals the change in momentum of the body. So
J = m(v–u)
J/m = v – u
V= J/m + u
Vx = Jx/m + ux
Vx = –1.14/0.057 + 20
Vx = -20 + 20 = 0m/s
Vx = 0m/s
Vy= Jy/m + uy
Vy= 0.33/0.057 + (-4.0)
Vy= 5.79 + (-4.0) = 1.79m/s
V = (0m/s)ı^−(1.79m/s)ȷ^
Answer:
This is the answer that I got.
Explanation:
Just check over it once.
